Chapter 6b.ppt

8/11/2019 Chapter 6b.ppt

1/35

1

BEX 17003

ELECTRICAL AND ELECTRONICSTECHNOLOGY

CHAPTER 6(b)


2/35

2

Chapter 6:SINUSOIDAL ALTERNATE CURRENT

AND VOLTAGE CIRCUITS


3/35

3

Lecture Contents1. Introduction2. Generating Sinusoidal Voltages3. The Sine Wave4. Phase Relationships5. Average and RMS Value


4/35

4

6.1 INTRODUCTION

Direct current (DC) is known as the electricity flowing in a constant

direction, and/or possessing a voltage with constant polarity.

DC is the kind of electricity made by a battery with definite positiveand negative terminals.

While, the alternating current (AC) is the sources that produce

voltages with alternating polarity, reversing positive and negative

over time.

In other words, the alternating current is the waveform where its

magnitude always varies with respect to time (emphasis is given to

the sine wave).


5/35

5

ALTERNATING CURRENT

Types of ac waveform:

Sinusoidal waveform Square waveform

Saw tooth waveformTriangle waveform


6/35

6

Advantages ac over dc

Many appliances need a large supply of current using ofac/dc generator AC generator larger, less complex internally, cheaper to operate

AC voltage easily and efficiently be transformed up/downusing transformer

method for dc voltages is inefficient ad more complex

Nearly all electronic circuits/equipment powered by dcvoltages

ac power arrives at home have to be converted into dc power

WHY ALTERNATING CURRENT?


7/35

7

6.2 GENERATING SINUSOIDAL VOLTAGESOne way to generate sinusoidal voltage is to rotate a coil of wire in apermanent magnetic field.

Figure below shows the simplified ac generator to generate ac voltage.

The magnitude of the resultant voltage is proportional to the rate at whichthe flux lines are cut.

Basic Generator Construction


8/35

8

360 0 GENERATOR OPERATION


9/35

9

GENERATING AC VOLTAGES

Coil voltage versus angularposition


10/35

10

WHAT TO KNOW??? Based-on Problem

Find the period and frequency of the waveform. Then, find rms current, rmsvoltage, Vp, Vpp and average voltage across R1. Voltage source is in rms value.

Vs

100 V

R1

R2

k 0.1

560

1. Period, T = ?

2. Frequency = ?

3. RMS Current, Irms = ?

4. RMS Voltage, Vrms = ?

5. Peak Voltage, Vp = ?

6. Peak to Peak Voltage, Vpp = ?

7. Average Voltage, Vavg = ?

Vs

0

V p

t5 ms


11/35


12/35

12

THE SINE WAVE cont..

SINE WAVECircuits driven bysinusoidal current orvoltage sources iscalled ac circuits .When a sinusoidalvoltage source (Vs) isapplied to a resistivecircuit, an alternatingsinusoidal currentresults.When the voltagechanges polarity, thecurrentcorrespondingly

changes direction.

R

+

-

V s

I

Positive alternation

R +

-

Vs

I

Negative alternation

(a) Positive voltage :current direction as shown

(b) Negative voltage :current reverses direction


13/35

13

Consider a sinusoidal wave in Figure A Thesinusoidal signal can be expressed as

where V m is the peak value (amplitude), is the angular frequency , is the phaseangle and T is the period.

Sinusoidal signals are periodic, repeating thesame pattern in interval T, which is

t cosV t v m

Figure A

t vT t v



14/35

14

The sine/cosine function completes 1 cycle in every 2radian, therefore

Frequency of sinusoidal signal is the number ofcompleted cycle in one second, which is

The angular frequency is given as

Angular frequency has a unit of radian/second (rad/s).

T

2 f 2

T f

1

2 T



15/35

15

Phase relationship: Assume there are two sinusoidalvoltage signal as shown in FigureB.Since v 2 (t) reaches the peak valuefirst, therefore it is said that v 2 (t) leads v 1 (t) by or v 1 (t) lags v 2 (t) by f .

In this situation, both v 1 and v 2 arealso said to be out of phase.If f = 0, then v 1 (t) and v 2 (t) are saidto be in phase , in which they reachthe minimum and maximum pointat the same time.

Figure B



16/35

16


MESUREMENT OF AC MAGNITUDEOne way to express the magnitude ( amplitude ), of an ACquantity is to measure its peak height on a waveform graph.

This is known as the peak value of an AC waveform


17/35

17


MESUREMENT OF AC MAGNITUDE Another way is to measure the total height between opposite

peaks. This is known as the peak-to-peak (P-P) value of an ACwaveform


18/35

18

THE SINE WAVE cont..The Period and Frequency of AC Waveform

A sine wave continues to repeat itself in identical cycles.The period ( T ) of sine wave is the time required to complete one full cycle.

Vs

-Vs

0 t

Period (T)

Period (T)

Perio d (T)

1st cycle 2nd cycle 3rd cycle


19/35

19

THE SINE WAVE cont..The Period and Frequency of AC Waveform

The frequency ( f ) of a waveform is the number of cycles that sinewave completes in one second.

Frequency is in unit Hertz (Hz).V

-V

0 t

1 s

2 cycles per second (2 Hz)

Lower frequency: fewer cycles per second

V

-V

0 t

1 s

5 cycles per second (5 Hz)

Higher frequency: more cycles per second


20/35

20

THE SINE WAVE cont..Relationship between Frequency and Period

The relationship between frequency ( f ) and period ( T) are as

follows:

There is a reciprocal relationship between f and T.The sine wave with longer period goes to fewer cycles in onesecond.

T f

1 f

T 1


21/35

21

THE SINE WAVE cont..Example 1:

1) The period of a certain sine wave is 10 ms. What is the frequency2) The frequency of a sine wave is 60 Hz. What is the period?

Solution

1)

2)

Hz xT f 100101011

3

ms f

T 7.166011


22/35

22


Related Problem1. A certain sine wave goes through four

cycles in 20 ms. What is the frequency?

2. If what is f ? sT 15


23/35

23

6.4 PHASE RELATIONSHIPSMATHEMATICAL FORMULA for SINE WAVE

General expression SinV y m

y

0

V m

Phase shifted (lagging)

y

0

V m

f

)( f SinV y my

0

V m

f

Phase shifted (leading)

)( f SinV y m


24/35

24

PHASE RELATIONSHIPS cont..

Example 2:Determine the equation for the waveform in the figure below if f = 60

Hz.

A I m 4

At Sint i )120377(2)( 0

i(t)

0 t

2

1200

-2

i(t)

0 t

4

500

-4

A I m

2usoid theof ument thet

sradiansin frequencyangular the f

sinarg

/377)60(22

)( f SinV y m At Sint i )40377(4)( 0

)( f SinV y m


25/35

25

THE COSINE WAVE

COSINE WAVE

General expression

CosV v m SinV v m

CosV v m


26/35

26

PEAK VALUE & PEAK-TO-PEAK VALUE

Several ways to express the value ofthe wave in terms of its voltage orcurrent magnitude.

Instantaneous Value, v The value at any point in time onsine/cosine wave.

Peak Value,___The value of voltage or current atthe positive or negative maximum(peak) wrt zero.

Peak-to-peak Value,____Is the voltage or current from thepositive peak to the negative peakof the waveform.

SinV v p CosV v p

V(V)

0 t

Vm

pp V p V

p V

p V

pp V


27/35

27

RMS or EFFECTIVE VALUE

AC waveform is only at its maximum value for an instant intime, spending most of its time between peak currents.

rms is also referred to as the effective value.The effective value of a sine wave is equal to 0.707 of thepeak value.

rms = 0.707 x peak

roo t m ean s qu are (rm s) resu lt of 0.707 can always tel lhow effect ive an ac s ine wave wi l l be .


28/35

28

RMS or EFFECTIVE VALUE cont..

Eg;

A dc source would be 10 A effective because it deliver powercontinuously.

A 10 A ac source would only be 7.07 A effective, because it is at 10 Afor only a short period of time.

Unless stated, ac value of voltage/current are always given in rms

Effective Equivalent


29/35

29

6.5 RMS and AVERAGE VALUE

The peak value of a sine wave can be converted to the corresponding r o o tm ean s qu are (rms ) or effec t ive value . The conversion process is as follows:

From the equation of sine wave,

SinV v p (5.1) Square the eq. (5.1)

222 SinV v p (5.2)

Obtain the mean or average value of

by dividing the area under a half-cycle

of the curve by .

The area is found by integration and

trigonometric identities.

2v


30/35

30

RMS and AVERAGE VALUE

(5.3)

2

02

2sin21

2

2cos12

sin1

2

2

0

2

0

2

2

0

2

2

p

p

p

p

p

avg

V

V

V

d V

d V

areaV

Finally

p p

avg rms V V

V V 707.02

2 (5.4)


31/35

31

The average value of the positive/negative alternation is found by taking eitherthe positive/negative alternation, and listing the amplitude or vector length ofcurrent or voltage at 1 0 intervals.

Sum of these values divided by total number of values (averaging).

average = 0.637 x peak

AVERAGE VALUE


32/35

32

AVERAGE VALUE

(5.5)

p

p

p

p

avg

V

V

V

d V

areaV

637.0

2

cos

sin1

0

0

The average value is the total area under the half-cycle curve divided by thedistance in radians of the curve.


33/35

33

REMEMBER !

p p

rms V V V 707.0

2 p pavg V V V 637.02 T

f 1 f

T 1

SinV v m

CosV v m


34/35

34

RMS and AVERAGE VALUE Example 3: Find the period and frequency of the waveform. Then, find rms current , rms

voltage , Vp, Vpp and average voltage across R1 . Voltage source is in rms value.

Vs100 V

R1

R2

k 0.1

560

Solution:

1. Period (T)

T = 10 ms

2. Frequency

f = 1/T = 100 Hz3. Total resistance

k

k RT

56.1

5600.1

4. Use Ohms law to find rms current

mAk

V

R

V I

T

rms

rms1.64

56.1

100

5. The rms voltage across R1

V R I V rmsrms 1.641)(1

6. Vp, Vpp and Vavg across R1

V V

V rms p 66.90707.0)(1

V xV pp 32.181)266.90(

V V V pavg 75.57637.0

Vs

0

V p

t5 ms


35/35

35

Example 4

V)1050cos(12)( t t v

Hz958.71

isfrequencyThe

s1257.05022

periodThe

rad/s50isfrequencyangularThe

10is phaseThe

V12VmisamplitudeThe

:Solution

T f

T

f

Find the amplitude, phase, period , and frequency of the sinusoid

Documents

Chapter 6b.ppt